1994-1962 = 32 years
A = Ao*2^-t/27, so
A/Ao = 2^-32/27 = .4397 = 43.97%
In the early 1960s, radioactive strontium-90 was released during atmospheric testing of nuclear weapons and got into the bones of people alive at the time. If the half-life of strontium-90 is 27 years, what fraction of the strontium-90 absorbed in 1962 remained in people's bones in 1994?
2 answers
1994-1962 = 32 years
(.5)^(1/27)=.974654
.5 because half life
.974654^32
Answer: 43.977
(.5)^(1/27)=.974654
.5 because half life
.974654^32
Answer: 43.977